Callan-Symanzik equation for Effective Potential

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thatboi
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Hey all,
I am looking equations (13.24),(13.25) in Peskin & Schroeder's QFT book and I am confused about how they change from the Callan-Symanzik equation for the Effective Action to the Effective Potential. I thought the relation for constant ##\phi_{cl}## was ##\Gamma[\phi_{cl}] = -(VT)\cdot V_{eff}(\phi_{cl})##, equation (11.50) in the book. But making such a substitution into (13.24), I do not understand how to get to (13.25).
Any advice would be appreciated, thanks.
 
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First imagine that the space is discretized, so that you can write
$$\int dx\, \phi(x)\frac{\delta}{\delta\phi(x)}\Gamma[\phi]=
\sum_{x=1,2,\cdots} \phi_x \frac{\partial}{\partial\phi_x} \Gamma(\phi_1,\phi_2,\ldots)$$
Then define the quantity
$$V(\phi)=\Gamma(\phi,\phi,\ldots)$$
Clearly
$$\frac{\partial V}{\partial\phi}=\sum_{x}\left(\frac{\partial \Gamma}{\partial\phi_x} \right)_{\phi_1=\phi_2=\cdots=\phi}$$
so
$$\phi\frac{\partial V}{\partial\phi}=
\sum_{x}\phi\left(\frac{\partial \Gamma}{\partial\phi_x} \right)_{\phi_1=\phi_2=\cdots=\phi} =
\left( \sum_{x} \phi_x\frac{\partial \Gamma}{\partial\phi_x} \right)_{\phi_1=\phi_2=\cdots=\phi}$$
Finally turn back to the continuous ##x##, so that the last formula can be written as
$$\left( \int dx\, \phi(x)\frac{\delta}{\delta\phi(x)}\Gamma[\phi]\right)_{\phi(x) =\phi, \; \forall x}
=\phi\frac{\partial V}{\partial\phi}$$
Now getting (13.25) should be obvious.
 
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